In Δ ABC, the perpendiculars drawn from A, B and C meet the opposite sides at points D, E and F, respectively. AD, BE and CF intersect at point P. If ∠ EPD = 110° and the bisectors of ∠ A and ∠ B meet at point Q, then ∠ AQB = ?

125°

As we know,

⇒ \(\angle\)C + \(\angle\)EPD = \({180}^\circ\)

⇒ \(\angle\)C = \({180}^\circ\) - \({110}^\circ\) = \({70}^\circ\)

Again, as we know

⇒ \(\angle\)AQB = \({90}^\circ\) + \(\angle\)C/2

⇒ \(\angle\)AQB = \({90}^\circ\) + 70/2 = \({90}^\circ\) + \({35}^\circ\) = \({125}^\circ\).